How do we know that electron wave function extends to infinity?

Question

Why do physicists assume this? Is it a proven fact that wave function extends to infinity or just a theory? Would it make sense if they didn't extend to infinity?

Answer 1

Is it a proven fact that wave function extends to infinity or just a theory?

The mathematical representations of the wavefunctions extends to infinity since there are no boundary conditions to limit the distance. It is just a theory since we cannot go to infinity to test the wavefunctions.

Would it make sense if they didn't extend to infinity?

According to the mathematical models, no.

Answer 2

First of all, remember we are talking physics here, and in physics zero and so small we cannot measure it are just the same thing.

Having said that, it is not true that one always assume that the "electron wave function extends to infinity". One of the most common models is the so-called particle in a box, in which we impose the particle (electron or whatever you like) to be confined in a given volume $V$. This means that its wave function is zero outside that volume $V$.

It turns out that if you do the math for a particle confined in a box of volume $V$, with $V$ very large, all the predictions are identical to those coming from the model in which we assume the electron wave functions to be infinitely extended.

This may sound strange, but it is due to the fact that even those "infinitely extended wave functions" are not really describing the objects you would measure in the real world. They are just used as a basis (just like a basis of a vector space), describing through superposition the "real" wave functions of the objects of interest.

If you want to describe an electron travelling through space in a way that resembles what you would expect from a particle in the classical sense, you have to use a wave packet, which through an appropriate superposition of (for example) infinitely extended wave functions realizes a finitely extended wave function travelling at a given velocity (as you can see for example here)

Answer to the comments:

It's true that the particle in a box has a finite wave function, but it's also not a real potential - it requires that there is an infinite force at two locations, which can't happen.

This is true, of course. That of an infinitely high potential barrier is an approximation of how things really are. However, there are two points to notice here:

1. It can be a very good approximation, or just a first approximation used to get a qualitative feeling of what's going on in a more complex system. Either way the formalism itself has no problem dealing with it, and just to repeat this point, physically infinitely high and very very high are the same thing.
2. Even if there are no "walls" or things actually confining the particles, we can always assume there is some very large volume $V$ from which the particles will never escape (we just take $V$ to be larger than any characteristic property of the system), thus using a particle in a box model. This is done mainly for formal reasons: for one, it allows to count the number of states and talk of density of states.

why is it that particles must be described by finitely extended functions?

It really depends of what you mean with "particle". In the classical sense, a particle is an object moving in space with a more or less defined position. To describe something like this in quantum mechanics you need a wave packet with a wave function non vanishing only in some finite region of space. Again, that wave function can be considered to abruptly vanish at some point making the particle really confined in some region, but this is mathematically very "unnatural" to describe. It is way easier to just take the wave function to be vanishingly small outside the given region because this is, as said above, physically indistinguishable from the former case.

Answer 3

It does not extend to infinity, that does not make sense. It is possible to have such a solution for a Schrödinger equation but it is not physical. You have to have a solution that is in the form of finite wave packets instead. And you can achieve this by linear superposition of many wave functions that have suitable form. If the wave packet was infinite then the probability density would be infinite too and this is unacceptable. That is why we use only square integrable functions when describing physical systems.

Answer 4

Take a Gaussian function as an example of a wave function. It extends to infinity, but the probability associated to finding the particle away from its mean position approaches zero rapidly, even though it is never strictly zero.

If all the universe was a stationary single electron, would you make any sense to impose any boundary to its wave function? If there was a limited-energy barrier, would it impose a strict boundary? The only case the wave function does not extend to infinity is when it is in a infinite potential well.

Answer 5

If you present a model of a finite electric field of the electron, one has to determine its effective diameter. Such a model could include charged particles (electron, proton, positron, ...) and photons that have the same elementary quanta. There has to be a negative and a postives quanta. Each electron at equal potential points will contain then the same number of quanta, and thus have the identical effective diameter.

There have to be two types of quanta in the model, they could built strings respectively field lines. For electrons or antiprotons at the ends of such field lines sit negative quantum and for protons or positrons at the ends sit positive quantum.

For the emission of photons an equal number of positive and negative quantum leaves the particle which is responsible for the emission. Photons are then composite particles and there is a finite number of energy levels within an given energy interval.

It's a model only.